Let $h(x)=e^{2x-6}-e$ Where does $h$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=\dfrac{5}{2}$ (Choice C) C $x=3$ (Choice D) D $h$ has no critical points.
A critical point of $h$ is a point in the domain of $h$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $h$, let's find its derivative. $\begin{aligned} h'(x)&=\dfrac{d}{dx}\left[ e^{2x-6}-e \right] \\\\ &=e^{2x-6} \cdot \dfrac{d}{dx}[2x-6] \\\\ &=2e^{2x-6} \end{aligned}$ Now let's look for $x$ -values where $h'$ is zero or undefined. $2e^{2x-6}=0$ has no solution, so $h'$ never equals $0$. $2e^{2x-6}$ is never undefined, so $h'$ is never undefined. In conclusion, $h$ has no critical points.